The mass is 224 grams of NaOH
As I understand from your question, we should synthesize nickel sulfate first from nickel (II) oxide and sulfuric acid and second from nickel carbonate and sulfuric acid.
The chemical reactions will look like this:
NiO (s) + H₂SO₄ (aq) → NiSO₄ (aq) + H₂O (l)
NiCO₃ (aq)* + H₂SO₄ → NiSO₄ (aq) + H₂CO₃ (aq)
but carbonic acid will decompose to carbon dioxide and water
H₂CO₃ (aq) → CO₂ (g) + H₂O (l)
(*) NiCO₃ has a poor solubility in water, but enough to start the reaction.
Answer is: D. Cl (chlorine).
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Barium, potassium and arsenic are metals (easily lost valence electrons), chlorine is nonmetal (easily gain electrons).
Alkaline metals (in this example, potassium) have lowest ionizations energy and easy remove valence electrons (one electron), earth alkaline metals (in this example, barium) have higher ionization energy than alkaline metals, because they have two valence electrons.
Nonmetals (in this example chlorine) are far right in the main group and they have highest ionization energy, because they have many valence electrons.
Your weight would be less but your mass would remain the same.
Answer:
a) kc= [SO3 ]/([SO2 ][O2 ])
b) kc= 2.27*10⁶ M⁻¹
v) the reaction is product-favored
Explanation:
for the reaction, the equilibrium constant is
SO2 (g) + O2 (g) <-----> SO3 (g)
he equilibrum constant is
kc= [SO3 ]/([SO2 ]*[O2 ])
replacing values
kc= [SO3 ]/([SO2 ]*[O2 ]) = 1.01*10⁻² M/(3.61*10⁻³M*6.11 x 10⁻⁴ M) = 2.27*10⁶ M⁻¹
since kc>>1 the reaction is product-favored
